# Field Product with Non-Zero Element yields Unique Solution

## Theorem

Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Let $a, b, x \in F$ such that $b \ne 0_F$.

Let:

$b \times x = a$

Then:

$x = a b^{-1}$

That is:

$x = \dfrac a b$

where $\dfrac a b$ denotes division.

## Proof

 $\ds b \times x$ $=$ $\ds a$ with $b \ne 0_F$ $\ds \leadsto \ \$ $\ds b^{-1} \times \paren {b \times x}$ $=$ $\ds b^{-1} \times a$ multiplying both sides by $b^{-1}$, which exists because $b \ne 0$ $\ds \leadsto \ \$ $\ds \paren {b^{-1} \times b} \times x$ $=$ $\ds b^{-1} \times a$ Field Axiom $\text M1$: Associativity of Product $\ds \leadsto \ \$ $\ds 1_F \times x$ $=$ $\ds b^{-1} \times a$ Field Axiom $\text M4$: Inverses for Product $\ds \leadsto \ \$ $\ds x$ $=$ $\ds b^{-1} \times a$ Field Axiom $\text M3$: Identity for Product

$\blacksquare$